Change of basis of polynomial mapping.

Question

My question is whether I could change the basis of polynomial $P^{13}$ meaning changing $\{1,x,x^2,\ldots,x^{13}\}$ to $\{x^2,\ldots,x^{13},x,1\}$?

Answer 1

First of all the sets are identical since a set is only concerned with whether an element is in the set or not (and not concerned with the order of the elements).

With basis one often just mean a set of independent vectors spanning the space.

However in some cases one want to talk about ordered basis where one take order into account. Such a basis should not be written using set notation, but rather use tuple notation:

$$(1,x,x^2,\ldots,x^{13})$$ vs $$(x^2,\ldots,x^{13},x,1)$$

By permuting the vectors in such a basis one will still end up with a basis since they still are linear independent and spans the same space. This has nothing to do with that the space is a space of polynomials.

To prove that we can consider an ordered base $e_j$ and a permutation $\Pi$ (a bijective mapping on the index set). This gives us a new base $f_j = e_{\Pi(j)}$.

Now we show that any vector $u$ can be written as a linear combination of $f_j$, but since $e_j$ is a base $u$ can be written in that base $u=\sum c_j e_j$, but since $\Pi$ is bijective it has an inverse and $e_j = f_{\Pi^{-1}(j)}$ so $u = \sum c_j f_{\Pi^{-1}(j)}$.

In similar way we can show that $f_j$ are independent. If $0 = \sum c_j f_j$ we have $0 = \sum c_{\Pi^{-1}(j)} e_{j}$ which means that $c_j=0$ since $e_j$ are independent.